Find the sum of G.P. :
x+yx−y+1+x−yx+y\dfrac{x + y}{x - y} + 1 + \dfrac{x - y}{x + y}x−yx+y+1+x+yx−y + …….. upto n terms
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Common ratio (r) = 1x+yx−y=x−yx+y\dfrac{1}{\dfrac{x + y}{x - y}} = \dfrac{x - y}{x + y}x−yx+y1=x+yx−y.
S=a(1−rn)(1−r)..........(As∣r∣<1)=x+yx−y[1−(x−yx+y)n]1−x−yx+y=(x+y)[1−(x−yx+y)n](x−y)x+y−x+yx+y=(x+y)2[1−(x−yx+y)n]2y(x−y)=(x+y)22y(x−y)[1−(x−yx+y)n].S = \dfrac{a(1 - r^n)}{(1 - r)} ……….(As |r| \lt 1) \\[1em] = \dfrac{\dfrac{x + y}{x - y}\Big[1 - \Big(\dfrac{x - y}{x + y}\Big)^n\Big]}{1 - \dfrac{x - y}{x + y}} \\[1em] = \dfrac{(x + y)\Big[1 - \Big(\dfrac{x - y}{x + y}\Big)^n\Big]}{(x - y)\dfrac{x + y - x + y}{x + y}} \\[1em] = \dfrac{(x + y)^2\Big[1 - \Big(\dfrac{x - y}{x + y}\Big)^n\Big]}{2y(x - y)} \\[1em] = \dfrac{(x + y)^2}{2y(x - y)}\Big[1 - \Big(\dfrac{x - y}{x + y}\Big)^n\Big].S=(1−r)a(1−rn)……….(As∣r∣<1)=1−x+yx−yx−yx+y[1−(x+yx−y)n]=(x−y)x+yx+y−x+y(x+y)[1−(x+yx−y)n]=2y(x−y)(x+y)2[1−(x+yx−y)n]=2y(x−y)(x+y)2[1−(x+yx−y)n].
Hence, sum = (x+y)22y(x−y)[1−(x−yx+y)n].\dfrac{(x + y)^2}{2y(x - y)}\Big[1 - \Big(\dfrac{x - y}{x + y}\Big)^n\Big].2y(x−y)(x+y)2[1−(x+yx−y)n].
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1−12+14−18+.......1 - \dfrac{1}{2} + \dfrac{1}{4} - \dfrac{1}{8} + …….1−21+41−81+……. to 9 terms
1−13+132−1331 - \dfrac{1}{3} + \dfrac{1}{3^2} - \dfrac{1}{3^3}1−31+321−331 + …… to n terms
3+13+133+.......\sqrt{3} + \dfrac{1}{\sqrt{3}} + \dfrac{1}{3\sqrt{3}} + …….3+31+331+……. to n terms
How many terms of the geometric progression 1 + 4 + 16 + 64 + …….. must be added to get sum equal to 5461 ?
